The sum of all integral values of $$p$$ such that the equation $$3\sin^2 x + 12\cos x - 3 = p, x \in R,$$ has at least one solution is :

To find the sum of all integral values of $$p$$, we first need to determine the range of the function $$f(x) = 3\sin^2 x + 12\cos x - 3$$.

1. Simplify the Function

Since we have both $$\sin^2 x$$ and $$\cos x$$, let's convert everything into terms of $$\cos x$$ using the identity $$\sin^2 x = 1 - \cos^2 x$$:

$$f(x) = 3(1 - \cos^2 x) + 12\cos x - 3$$

$$f(x) = 3 - 3\cos^2 x + 12\cos x - 3$$

$$f(x) = -3\cos^2 x + 12\cos x$$

2. Find the Range

Let $$t = \cos x$$. Since $$x \in R$$, we know that $$-1 \le t \le 1$$.

Now the function becomes a quadratic in terms of $$t$$:

$$g(t) = -3t^2 + 12t, \quad t \in [-1, 1]$$

To find the maximum and minimum values of $$g(t)$$ on the interval $$[-1, 1]$$:

• At $$t = -1$$: $$g(-1) = -3(-1)^2 + 12(-1) = -3 - 12 = -15$$
• At $$t = 1$$: $$g(1) = -3(1)^2 + 12(1) = -3 + 12 = 9$$
• Vertex check: The vertex of a quadratic $$at^2 + bt + c$$ is at $$t = -b/2a$$.
• First term ($$a$$) = $$-15$$
• Last term ($$l$$) = $$9$$
• Number of terms ($$n$$) = $$9 - (-15) + 1 = 25$$

$$t = \frac{-12}{2(-3)} = \frac{-12}{-6} = 2$$

Since $$t=2$$ is outside our interval $$[-1, 1]$$, the function is monotonic (specifically, strictly increasing) within the range of $$t$$.

Therefore, the range of $$p$$ is $$[-15, 9]$$.

3. Calculate the Sum of Integral Values

The integral values of $$p$$ are: $$-15, -14, -13, \dots, 0, \dots, 8, 9$$.

We can use the arithmetic progression sum formula $$S = \frac{n}{2}(\text{first term} + \text{last term})$$:

$$\text{Sum} = \frac{25}{2}(-15 + 9)$$

$$\text{Sum} = \frac{25}{2}(-6)$$

$$\text{Sum} = 25 \times (-3) = -75$$

Correct Answer: C (-75)